11.1.3: Products and Quotients Raised to Powers
- Page ID
- 67617
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- Raise a quotient to a power.
- Simplify expressions using a combination of the properties.
Introduction
The rules of exponents are very useful when simplifying and evaluating expressions. When multiplying, dividing, or raising a power to a power, using the rules for exponents helps to make the process more efficient. Now let’s look at rules for taking a product or a quotient to a power.
A Product Raised to a Power
Once the rules of exponents are understood, you can begin solving more complicated expressions more easily. Recall that when you take a power to a power, you multiply the exponents, \(\ \left(x^{a}\right)^{b}=x^{a \cdot b}\).
What happens when you raise an entire expression inside parentheses to a power? You can use the techniques you already know to simplify this expression.
\(\ (2 a)^{4}=(2 a)(2 a)(2 a)(2 a)=(2 \cdot 2 \cdot 2 \cdot 2)(a \cdot a \cdot a \cdot a)=\left(2^{4}\right)\left(a^{4}\right)=16 a^{4}\)
Notice that the exponent is applied to each factor of \(\ 2a\). So, we can eliminate the middle steps.
\(\ \begin{array}{l}
(2 a)^{4}&=\left(2^{4}\right)\left(a^{4}\right) \text {, applying the } 4 \text { to each factor, } 2 \text { and } a\\
&=16 a^{4}
\end{array}\)
The product of two or more numbers raised to a power is equal to the product of each number raised to the same power.
For any nonzero numbers \(\ a\) and \(\ b\) and any integer \(\ x\), \(\ (a b)^{x}=a^{x} \cdot b^{x}\)
Caution! Do not try to apply this rule to sums. Think about the expression \(\ (2+3)^{2}\). Does \(\ (2+3)^{2}\) equal \(\ 2^{2}+3^{2}\)? No, it does not. \(\ (2+3)^{2}=5^{2}=25\) and \(\ 2^{2}+3^{2}=4+9=13\). So, you can only use this rule when the numbers inside the parentheses are being multiplied (or divided, as we will see next).
Simplify. \(\ (2 y z)^{6}\)
Solution
| \(\ 2^{6} y^{6} z^{6}\) | Apply the exponent to each number in the product. |
\(\ (2 y z)^{6}=64 y^{6} z^{6}\)
If the variable has an exponent with it, use the Power Rule: multiply the exponents.
Simplify. \(\ \left(-7 a^{4} b\right)^{2}\)
Solution
| \(\ (-7)^{2}\left(a^{4}\right)^{2}(b)^{2}\) | Apply the exponent 2 to each factor within the parentheses. |
| \(\ 49 a^{4 \cdot 2} b^{2}\) | Square the coefficient and use the Power Rule to square \(\ \left(a^{4}\right)^{2}\). |
| \(\ 49 a^{8} b^{2}\) | Simplify. |
\(\ \left(-7 a^{4} b\right)^{2}=49 a^{8} b^{2}\)
Simplify the expression. \(\ \left(-3 x^{3} y^{2}\right)^{4}\)
- \(\ x^{4} y^{4}\)
- \(\ -81 x y^{8}\)
- \(\ 81 x^{12} y^{8}\)
- \(\ 81 x^{7} y^{6}\)
- Answer
-
- Incorrect. Remember to apply the exponent to all terms within the parentheses, including the coefficient. \(\ \left(-3 x^{3} y^{2}\right)^{4}=(-3)^{4}\left(x^{3}\right)^{4}\left(y^{2}\right)^{4}\); the correct answer is \(\ 81 x^{12} y^{8}\).
- Incorrect. Remember to apply the exponent to all terms within the parentheses, including the variable \(\ x\). \(\ \left(-3 x^{3} y^{2}\right)^{4}=(-3)^{4}\left(x^{3}\right)^{4}\left(y^{2}\right)^{4}\); the correct answer is \(\ 81 x^{12} y^{8}\).
- Correct. \(\ \left(3 x^{3} y^{2}\right)^{4}=(-3)^{4}\left(x^{3}\right)^{4}\left(y^{2}\right)^{4}=81 x^{12} y^{8}\).
- Incorrect. Remember to multiply exponents (not add them) when you are raising a power to a power. In this case, \(\ \left(-3 x^{3} y^{2}\right)^{4}=(-3)^{4}\left(x^{3}\right)^{4}\left(y^{2}\right)^{4}=(-3)^{4} \cdot x^{3 \cdot 4} \cdot y^{2 \cdot 4}\). The correct answer is \(\ 81 x^{12} y^{8}\).
A Quotient Raised to a Power
Now let’s look at what happens if you raise a quotient to a power. Suppose you have \(\ \frac{3}{4}\) and raise it to the 3rd power.
\(\ \left(\frac{3}{4}\right)^{3}=\left(\frac{3}{4}\right)\left(\frac{3}{4}\right)\left(\frac{3}{4}\right)=\frac{3 \cdot 3 \cdot 3}{4 \cdot 4 \cdot 4}=\frac{3^{3}}{4^{3}}\)
You can see that raising the quotient to the power of 3 can also be written as the numerator (3) to the power of 3, and the denominator (4) to the power of 3.
Similarly, if you are using variables, the quotient raised to a power is equal to the numerator raised to the power over the denominator raised to power.
\(\ \left(\frac{a}{b}\right)^{4}=\left(\frac{a}{b}\right)\left(\frac{a}{b}\right)\left(\frac{a}{b}\right)\left(\frac{a}{b}\right)=\frac{a \cdot a \cdot a \cdot a}{b \cdot b \cdot b \cdot b}=\frac{a^{4}}{b^{4}}\)
When a quotient is raised to a power, you can apply the power to the numerator and denominator individually, as shown below.
\(\ \left(\frac{a}{b}\right)^{4}=\frac{a^{4}}{b^{4}}\)
For any number \(\ a\), any non-zero number \(\ b\), and any integer \(\ x\), \(\ \left(\frac{a}{b}\right)^{x}=\frac{a^{x}}{b^{x}}\).
Simplify. \(\ \left(\frac{2 x y}{x}\right)^{3}\)
Solution
| \(\ \frac{2^{3} x^{3} y^{3}}{x^{3}}\) | Apply the power to each factor individually. |
| \(\ 2^{3} \cdot \frac{x^{3}}{x^{3}} \cdot \frac{y^{3}}{1}\) | Separate into numerical and variable factors. |
| \(\ 8 \cdot x^{(3-3)} \cdot y^{3}\) | Simplify by taking 2 to the third power and applying the Quotient Rule for exponents: subtract the exponents of matching variables. |
| \(\ 8 x^{0} y^{3}\) | Simplify. Remember that \(\ x^{0}=1\). |
\(\ \left(\frac{2 x y}{x}\right)^{3}=8 y^{3}\)
Simplify the expression. \(\ \frac{(a b c)^{2}}{(a b)^{2}}\)
- 1
- \(\ c^{2}\)
- \(\ (a b c)^{2}(a b)^{-2}\)
- \(\ \frac{a^{2} b^{2} c^{2}}{a^{2} b^{2}}\)
- Answer
-
- Incorrect. Follow the order of operations. Apply the exponent to each variable in the product first, and then use the Quotient Rule: subtract the exponents in terms that have the same base. The correct answer is \(\ c^{2}\).
- Correct. Apply the exponent to each variable in the product and then use the Quotient Rule: subtract the exponents in terms that have the same base.
- Incorrect. Although \(\ (a b c)^{2}(a b)^{-2}\) is equivalent to \(\ \frac{(a b c)^{2}}{(a b)^{2}}\), it is not in simplest form. Apply the exponent to each variable in the product and then use the Quotient Rule: subtract the exponents in terms that have the same base. The correct answer is \(\ c^{2}\).
- Incorrect. This is an equivalent expression but not in simplest form. Use the Quotient Rule: subtract the exponents in terms that have the same base. The correct answer is \(\ c^{2}\).
Simplifying and Evaluating Expressions with Exponents
Now let’s look at some more complex expressions and see how all the rules can help make working with exponents easier. While these expressions may look complicated at first, just keep following the rules of exponents to simplify them!
Simplify. \(\ \left(4 x^{3}\right)^{5} \cdot\left(2 x^{2}\right)^{-4}\)
Solution
| \(\ \frac{\left(4 x^{3}\right)^{5}}{\left(2 x^{2}\right)^{4}}\) | Recall that with negative exponents, you can use the reciprocal, and move it to the denominator with a positive exponent. |
| \(\ \begin{array}{l} \frac{4^{5}\left(x^{3 \cdot 5}\right)}{2^{4}\left(x^{2 \cdot 4}\right)} \\ \frac{1,024 x^{15}}{16 x^{8}} \end{array}\) |
Separate into numerical and variable factors and apply the exponent to each term. |
| \(\ 64 x^{15-8}\) | Divide 1,024 by 16, and subtract exponents using the Quotient Rule. |
\(\ 64 x^{7}\)
By using the rules of exponents, you’ve now got an equivalent expression, \(\ 64 x^{7}\), which is much easier to work with than \(\ \left(4 x^{3}\right)^{5} \cdot\left(2 x^{2}\right)^{-4}\).
The rules of exponents are also helpful when evaluating expressions for a certain value of a variable.
Evaluate \(\ \frac{24 x^{8} y^{2}}{\left(2 x^{3} y\right)^{2}}\) when \(\ x=4\) and \(\ y=-2\).
Solution
| \(\ \frac{24 x^{8} y^{2}}{2^{2} x^{3 \cdot 2} y^{2}}\) | In the denominator, notice that a product is being raised to a power. Use the rules of exponents to simplify the denominator. |
| \(\ \frac{24 x^{8} y^{2}}{4 x^{6} y^{2}}\) | Simplify \(\ 2^{2}\) and multiply the exponents of \(\ x\). |
| \(\ \left(\frac{24}{4}\right)\left(\frac{x^{8}}{x^{6}}\right)\left(\frac{y^{2}}{y^{2}}\right)\) | Separate into numerical and variable factors. |
| \(\ 6\left(x^{8-6}\right)\left(y^{2-2}\right)\) | Divide coefficients, use the Quotient Rule to divide the variables: subtract the exponents. |
| \(\ 6 x^{2} y^{0}=6 x^{2}\) | Simplify. Remember that \(\ y^{0}\) is 1. |
| \(\ \text { (6) }\left(4^{2}\right)=6 \cdot 16\) | Substitute the value 4 for the variable \(\ x\). |
\(\ \frac{24 x^{8} y^{2}}{\left(2 x^{3} y\right)^{2}}=96\) when \(\ x=4\) and \(\ y=-2\)
Notice that you could have worked this problem by substituting 4 for x and 2 for y in the original expression. You would still get the answer of 96, but the computation would be much more complex. Notice that you didn’t even need to use the value of \(\ y\) to evaluate the above expression.
Simplify. \(\ \left(3 a^{2} b^{3}\right)^{2}(-5 a b)\)
- \(\ \left(9 a^{4} b^{6}\right)(-5 a b)\)
- \(\ -15 a^{5} b^{7}\)
- \(\ -45 a^{5} b^{6}\)
- \(\ -45 a^{5} b^{7}\)
- Answer
-
- Incorrect. The first term has been raised to a power, but you can simplify further. The two factors haven't been multiplied together. The correct answer is \(\ -45 a^{5} b^{7}\).
- Incorrect. The coefficient of the first term must also be raised to a power. The correct answer is \(\ -45 a^{5} b^{7}\).
- Incorrect. When you take a power of a power, the exponents are multiplied. The correct answer is \(\ -45 a^{5} b^{7}\).
- Correct. In your answer, each variable appears only once, coefficients have been multiplied, and there are no powers of powers. \(\ \left(3 a^{2} b^{3}\right)^{2}(-5 a b)=\left(3^{2} a^{2 \cdot 2} b^{3 \cdot 2}\right)(-5 a b)=\left(9 a^{4} b^{6}\right)(-5 a b)\), which simplifies to \(\ -45 a^{5} b^{7}\).
Summary
There are many rules to use when working with exponential expressions. You can use these rules, as well as the definition of exponents, to simplify complex expressions. Simplifying an expression before evaluating the expression can often make the computation easier.